Physics, 07.12.2019 04:31 S145571

Find the mass m, the moment my, and the x-coordinate x of the center of mass of the uniform flat plate of density rho = 1 bounded by the curve y = 1 x2 + 6x + 13 , the line x = 2, and the x and y axes.

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Answer from: treaustin656

mass, m=0.4525

moment, M = 0.4222

x-coordinate = 0.9330

Explanation:

lets write out the useful formulas. $\\$

the $mass, m = \int\limits \,\int\limits {p} \, dA$ $\\$

where [tex dA=dxdy[/tex] $\\$

hence we can express the mass inters of x and y axes $\\$

$mass,m=\int\limits \int\limits {p} \,dx \, dy$ $\\$

Next we find expression for the moment $\\$

$moment ,M_{y}=mx = \int\limits \int\limits x{p} \,dx \,dy$

$\\$

lastly, the x-coordinate

$\\$

$x=\frac{moment,M_{y} }{mass,m}$

$\\$

Now from the question we were given the expression for y as

$\\$

$y= \frac{1}{\sqrt{x^{2}+6x+13 }}$

$\\$

and the points of x which are

$\\$

x=2 from 0. $\\$

also the densit, p=1

from the expression for the mass $\\$

$mass,m=\int\limits \int\limits {p} \,dx \,dy$ $\\$

$mass, m =\int\limits^2_0 \int\limits^y_0 {1} \,dx \,dy \\$

<img src= [\int\limits^y_0 \,dy ]\\" alt="mass.m=\int\limits^2_0 \, dx[y]^{y} _{0} \\" /> [\int\limits^y_0 \,dy ]\\" />

$mass.m=\int\limits^2_0 \,ydx$ $\\$

substitute the value of y $\\$

$mass,m=\int\limits^2_0 \,\frac{1}{\sqrt{x^{2}+6x+13}} dx\\$

we can manipulate the expression under the square-root to something familiar under integration $\\$

$(x^{2}+6x)+9+4 \\(x+3)^{2}+4\\$

Now we can easily integrate the expression using the integral of the inverse of an hyperbolic function. the identity is shown below $\\$

$\int\limits^a_b {\frac{1}{\sqrt{(u+a)^{2}+c^{2} } } } \, dx = sinh^{-1} \frac{u+a}{c} +k\\k = constant$

Hence

$mass,m=[sinh^{-1} (\frac{x+3}{2} )]^{2} _{0} \\$

$mass,m=[sinh^{-1} (\frac{2+3}{2} )]-[sinh^{-1} (\frac{0+3}{2} )]\\$

Note: $sinh^{-1}x \neq (sinhx)^{-1}=\frac{1}{sinhx} \\$

Rather we use the expansion $sinh^{-1}x=ln(x+\sqrt{1+x^{2} })\\$

$mass,m=1.6473-1.19483\\mass,m=0.4525\\$

Next we solve for the moment about the y-axis

from the expression for moment ,we have

$moment ,M_{y} =mx = \int\limits \int\limits x{p} \, dx \, dy$

$\\$

By following the same method of integral for the expansion process like we did above, we arrive at

$moment ,M_{y} =\int\limits^2_0 {\frac{x}{\sqrt{(x+3)^{2} +4 } } } \, dx\\$

integrating at the definite points,we have

$moment ,M_{y}=[\sqrt{(2+3)^{2}+4} -3sinh^{-1} (\frac{2+3}{2} )] - [\sqrt{(0+3)^{2}+4} -sinh^{-1} (\frac{0+3}{2} )]\\$

$moment ,M_{y} =0.4433-0.02106\\moment ,M_{y} =0.4222\\$

Finally we solve for the x-coordinate, using the ratio of the moment to the mass we obtained earlier

$x=\frac{moment}{mass}\\ x=\frac{0.4222}{0.4525} \\x=0.9330\\$

Answer from: Quest

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

Answer from: Quest

By dividing you can find the answer